An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.

Explanation

Since, total number of terms (n) `= 37 " " ` [odd]
` :.` Middle term `=((37+1)/(2))`th term = 19 th term
So, the three middle most terms = 18th, 19th and 20th.
By given condition,
Sum of the three middle most terms = 225
`a_(18)+a_(19)+a_(20) = 225`
`implies (a+17d)+(a+18d)+(a+19d) =225`
`implies 3a+54d=225`
` implies a +18d=75 " "` ...(i)
and sum of the last three terms = 429
`a_(35)+a_(36)+a_(37) = 429`
`implies (a+34d)+(a+35d)+(a+36d) =429`
`implies 3a+105d=429`
` implies a +35d=143 " "` ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
`17d=68`
`implies d=4`
From Eq. (i), `a+18(4)=75`
`implies a=75-72`
`implies a=3`
` :. ` Required AP is `a, a+d, a+2d, a+3d, ...`
i.e., `3, 3+4, 3+2(4), 3+3(4), ...`
i.e., `3, 7, 3+8, 3+12, ...`
i.e., `3, 7, 11, 15, ...`